Difference between revisions of "2012 AIME I Problems/Problem 6"
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Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{143} = z.</math> We know that <math>z \neq 0,</math> because we are given the imaginary part of <math>z,</math> so we can divide by <math>z</math> to get <math>z^{142} = 1.</math> So, <math>z</math> must be a <math>142</math>nd root of unity, and thus, by De Moivre's theorem, the imaginary part of <math>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.</math> Thus, <math>n = \boxed{071}.</math> | Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{143} = z.</math> We know that <math>z \neq 0,</math> because we are given the imaginary part of <math>z,</math> so we can divide by <math>z</math> to get <math>z^{142} = 1.</math> So, <math>z</math> must be a <math>142</math>nd root of unity, and thus, by De Moivre's theorem, the imaginary part of <math>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.</math> Thus, <math>n = \boxed{071}.</math> | ||
| − | ==Video | + | ==Video Solutions== |
https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s | https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s | ||
| + | |||
| + | https://www.youtube.com/watch?v=DMka35X-3WI&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=6 (Solution by Richard Rusczyk) - AMBRIGGS | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=5|num-a=7}} | {{AIME box|year=2012|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:15, 30 July 2022
Contents
Problem
The complex numbers 
and 
satisfy 
and the imaginary part of is 
, for relatively prime positive integers 
and 
with 
Find 
Solution
Substituting the first equation into the second, we find that 
and thus 
We know that 
because we are given the imaginary part of 
so we can divide by to get 
So, must be a 
nd root of unity, and thus, by De Moivre's theorem, the imaginary part of will be of the form 
where 
Note that 
is prime and 
by the conditions of the problem, so the denominator in the argument of this value will always be 
Thus, 
Video Solutions
https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
https://www.youtube.com/watch?v=DMka35X-3WI&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=6 (Solution by Richard Rusczyk) - AMBRIGGS
See also
| 2012 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
